Zero Order Reaction - Practice Questions & MCQ

Zero order Reactions

In such reactions rate of reaction is independent of concentration of the reactants.

        Rate $\propto[\text { concentration }]^0$
 For example, suppose we have a reaction

$A \longrightarrow B$

then, the rate of reaction can be written as 

Rate $=-\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0$

From the above equation it is evident that for a Zero order reaction,

(1) The rate of reaction is equal to the rate constant

(2) The rate of reaction is constant and independant of time

(3) The unit of rate constant is $\mathrm{molL}^{-1}$ time $^{-1}$

(4) The rate of reaction cannot be changed by changing the concentration of reactant.

Integrated Rate law for a Zero Order reaction

Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction,

$\mathrm{A} \rightarrow \mathrm{P}$

Rate $=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0$

$\Rightarrow$ Rate $=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}$

$\Rightarrow \mathrm{d}[\mathrm{A}]=-\mathrm{kdt}$

$\Rightarrow \int_{\left[\mathrm{A}_0\right]}^{\left[\mathrm{A}_{\mathrm{t}}\right]} \mathrm{d}[\mathrm{A}]=-\mathrm{k} \int_0^{\mathrm{t}} \mathrm{dt}$
Thus, on integrating both sides, we get:

$\left[\mathrm{A}_{\mathrm{t}}\right]=[\mathrm{A}]_0-\mathrm{kt}$

Comparing the above equation with the equation of a straight line, y = mx + c, if we plot [A] against t, we get a straight line as shown in the above figure with slope = –k and intercept equal to [A]_o.

Half-life of reaction: 

The half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its initial concentration. It is represented as t_1/2.
For a zero order reaction, rate constant is given as follows:

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0-\mathrm{kt}$
When $\mathrm{t}=\mathrm{t}_{\frac{1}{2}},[\mathrm{~A}]_{\mathrm{t}}=\frac{[\mathrm{A}]_0}{2}$
Putting these values in the integrated rate expression,
$\frac{[\mathrm{A}]_0}{2}=[\mathrm{A}]_0-\mathrm{kt}_{\frac{1}{2}}$

Upon solving the above expression we have, 
$\mathrm{t}_{\frac{1}{2}}=\frac{[\mathrm{A}]_0}{2 \mathrm{k}}$
Thus, it is clear that half life for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.

Life time of Reaction: It is time in which 100% of the reaction completes. It is represented as t_LF.
Thus, at t = t_LF, A = 0
Now, from integrated rate equation for zero order, we know:

$\mathrm{A}=\mathrm{A}_{\mathrm{o}}-\mathrm{kt}$

$0=\mathrm{A}_{\mathrm{o}}-\mathrm{kt}_{\mathrm{LF}}$
Thus, $\mathrm{t}_{\mathrm{LF}}=\frac{\mathrm{A}_0}{\mathrm{k}}$

This concept can be understood by the following example:

Example: 
The reaction occurs as follows:
$3 \mathrm{~A} \rightarrow \mathrm{P}$
Initial concentration of A is given as and rate of the reaction(r) is given as k[A]^o. Find the concentration of A after time 't' and also determine the half life of A.

Solution:
For zero-order reaction, the rate equation is given as follows:

$\mathrm{A}=\mathrm{A}_{\mathrm{o}}-\mathrm{kt}$

According to question, we have given:

$\mathrm{r}=\mathrm{k}[\mathrm{A}]^0=-\frac{1}{3} \frac{\mathrm{dA}}{\mathrm{dt}}$

$-\frac{1}{3} \frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{k} \quad\left(\right.$ Since $\left.[\mathrm{A}]^0=1\right)$

Now, integrating both sides we get:

$[\mathrm{A}]_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}}=-3 \mathrm{k}(\mathrm{t})_{\mathrm{o}}^{\mathrm{t}}$

$\mathrm{A}-\mathrm{A}_{\mathrm{o}}=-3 \mathrm{kt}$

$\mathbf{A}=\mathbf{A}_{\mathbf{o}}-3 \mathbf{k} \mathbf{t}$
This is the concentration of A after time 't'.

Now, for zero-order reaction, the half-life of a reaction is given as below:

$\mathrm{t}_{1 / 2}=\frac{\mathrm{A}_{\mathrm{o}}}{2 \mathrm{k}}=\frac{\mathrm{A}_{\mathrm{o}}}{2(3 \mathrm{k})}$

Thus, $\mathbf{t}_{1 / 2}=\frac{\mathrm{A}_{\mathrm{o}}}{6 \mathrm{k}}$
This is the half-life of A for this reaction.